2024 AMC 12A Problem 11

Here $2024_b=2b^3+2b+4$ $=2(b^3+b+2),$ so $2024_b$ is divisible by $16$ exactly when $b^3+b+2$ is divisible by $8.$ Checking residues $\bmod 8,$ $b^3+b+2\equiv0$ precisely for $b\equiv3,6,7\pmod8.$ Counting $b$ in $[5,2024]:$ residue $3$ gives $11,\ldots,2019$ ($252$ values), residue $6$ gives $6,\ldots,2022$ ($253$ values), and residue $7$ gives $7,\ldots,2023$ ($253$ values). So $K=252+253+253=758,$ and its digit sum is $7+5+8=20.$ Thus, the correct answer is D.